+118723 Hi, I think ths is what you are after :))
\(\frac{13x+2}{(x-1)^2+x+1}\\ =\frac{13x+2}{x^2-2x+1+x+1}\\ =\frac{13x+2}{x^2-x+2}\\ =\frac{13x+2}{(x-2)(x+1)}\\ =\frac{A}{(x-2)}+\frac{B}{(x+1)}\qquad Where\;\;A \;and\;B\; are\; real\;numbers.\\ =\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \)
\(\qquad so\\ \qquad Ax+A+Bx-2B=13x+2\\ \qquad (A+B)x+(A-2B)=13X+2\\ \qquad A+B=13\;\;(1) \qquad A-2B=2\;\;(2)\\ \qquad (1)-(2)\\ \qquad 3B=11\\ \qquad B=11/3\\ \qquad Substitution\;gives\; A=28/3\\ \)
\(=\frac{28}{3(x-2)}+\frac{11}{3(x+1)}\)
+118723 Hi, I think ths is what you are after :))
\(\frac{13x+2}{(x-1)^2+x+1}\\ =\frac{13x+2}{x^2-2x+1+x+1}\\ =\frac{13x+2}{x^2-x+2}\\ =\frac{13x+2}{(x-2)(x+1)}\\ =\frac{A}{(x-2)}+\frac{B}{(x+1)}\qquad Where\;\;A \;and\;B\; are\; real\;numbers.\\ =\frac{A(x+1)+B(x-2)}{(x-2)(x+1)}\\ \)
\(\qquad so\\ \qquad Ax+A+Bx-2B=13x+2\\ \qquad (A+B)x+(A-2B)=13X+2\\ \qquad A+B=13\;\;(1) \qquad A-2B=2\;\;(2)\\ \qquad (1)-(2)\\ \qquad 3B=11\\ \qquad B=11/3\\ \qquad Substitution\;gives\; A=28/3\\ \)
\(=\frac{28}{3(x-2)}+\frac{11}{3(x+1)}\)
